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Andrej [43]
3 years ago
7

Which graph represents a degree three polynomial function with a negative leading coefficient?

Mathematics
1 answer:
Vadim26 [7]3 years ago
7 0

The graph of a 5th-degree polynomial with a negative leading coefficient will start at (-∞, +∞) and end at (+∞, -∞). That is, it will have the overall shape \, downward from left to right. Along the way, it may change direction 0, 2, or 4 times, and may intersect the x-axis up to 5 times.

Only one of the graphs shown has the described end behavior. All of the others have the general shapes ...

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Is (8,8) a solution to the equation y=x?​
Nookie1986 [14]

Answer:

Yes

Step-by-step explanation:

(8,8) is a solution is y=x.

If we substitute the point into the equation, the equation is true.

(8,8) is in (x,y) format

Substitute:

y=x

8=8

LS=RS (left side equal rights side)

7 0
3 years ago
Read 2 more answers
In ΔMNO, the measure of ∠O=90°, the measure of ∠M=64°, and NO = 70 feet. Find the length of MN to the nearest tenth of a foot.
olga_2 [115]

Answer:

in triangle MNO

relationship between perpendicular and hypotenuse is given by sin angle

sin64°=p/h=70/x

x=70/sin64°=77.88foot

length of MN to the nearest tenth of a foot.

=78ft

6 0
3 years ago
Solve the following system of equations. Write each of your answers as a fraction reduced to lowest terms. In other words, write
morpeh [17]

Answer:

x = 57/28

y = -95/84

z = 97/168

Step-by-step explanation:

Use the application in the next link: https://www.zweigmedia.com/RealWorld/tutorialsf1/scriptpivotold.html

Start with the expanded array:

\left[\begin{array}{cccc}1&5&8&1\\3&2&2&5\\-2&-7&2&5\\\end{array}\right]

then using the tool provided, make row operations until you find the solution:

r2 = r2-3r1

\left[\begin{array}{cccc}1&5&8&1\\0&-13&-22&2\\-2&-7&2&5\\\end{array}\right]

r3 = r3+2r1

\left[\begin{array}{cccc}1&5&8&1\\0&-13&-22&2\\0&3&18&7\\\end{array}\right]

r2 = r2*(-1/13)

\left[\begin{array}{cccc}1&5&8&1\\0&1&22/13&-2/13\\0&3&18&7\\\end{array}\right]

r1 = r1- r2*5

\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&22/13&-2/13\\0&3&18&7\\\end{array}\right]

r3 = r3+ r2*-3

\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&22/13&-2/13\\0&0&168/13&97/13\\\end{array}\right]

r3 = r3*13/168

\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&22/13&-2/13\\0&0&1&97/168\\\end{array}\right]

r2 = r2- r3*22/13

\left[\begin{array}{cccc}1&0&-6/13&23/13\\0&1&0&-95/84\\0&0&1&97/168\\\end{array}\right]

r2 = r2+ r3*6/13

\left[\begin{array}{cccc}1&0&0&57/28\\0&1&0&-95/84\\0&0&1&97/168\\\end{array}\right]

Here you have a reduced array an therefore the answers to each variable are on each row:

\left[\begin{array}{c}x\\y\\z\end{array}\right]

8 0
3 years ago
Is ab perpendicular to cd ? Explain.
Afina-wow [57]
We know that
if two lines are perpendicular
then
the slopes 
m1*m2=-1

step 1
find the slope AB
A (0,2)
B (-3,-3)
m=(y2-y1)/(x2-x1)-----> m=(-3-2)/(-3-0)-----> m=-5/-3----> m1=5/3

step 2
find the slope CD
C (-4,1)
D (0,-2)
m=(y2-y1)/(x2-x1)-----> m=(-2-1)/(0+4)-----> m=--3/4----> m2=-3/4

step 3
multiply mi*m2
(5/3)*(-3/4)-----> -15/12
so
15/12 is not -1
therefore
 AB is not perpendicular to CD
4 0
4 years ago
What is the sum of the measures of interior angles of a 15-sides polygon
oksano4ka [1.4K]
Using
sum = (n-2)180
= (15-2)180 = 13×180 = 2340°
5 0
3 years ago
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